The well know Mumford-Shah functional functional $$ F(u)=\int_\Omega|\nabla u|^2+\mathcal H^{N-1}(S_u) \tag 1 $$ where $u\in SBV(\Omega)$ and $\nabla u$ is the absolutely continuous part of distribution derivative $Du$ and $S_u$ is the jump set of $u$.

The Functional in $(1)$ can be approximated by Sobolev function via Gamma convergence sense by following functional $$ F_\epsilon(u,v):=\int_\Omega \left[v^2|{\nabla u}|^2\right]dx +\int_{\Omega}\left[\epsilon|{\nabla v}|^2+\frac{1}{4\epsilon}{(v-1)^2}\right]dx \tag 2 $$

The original proof is really long.

My question: given that $u\in SBV\cap L^\infty$, can we interchange the energy part in $(1)$ and $(2)$ from $|\nabla u|^2$ to $|\nabla u|$? i.e., form superlinear to linear growth? i.e., if I define $$ E(u)=\int_\Omega|\nabla u|+\mathcal H^{N-1}(S_u) \tag 3 $$

$$ E_\epsilon(u,v):=\int_\Omega \left[v|{\nabla u}|\right]dx +\int_{\Omega}\left[\epsilon|{\nabla v}|^2+\frac{1}{4\epsilon}{(v-1)^2}\right]dx \tag 4 $$ may I have $E_\epsilon$ gamma convergence to $E$?

I understand there is a difference between super-linear growth and linear growth, but for my opinion that $|\nabla u|$ is more suitable in term of $SBV$ norm.

Thank you!

update: I guess is that the lack of weak closeness of $W^{1,1}$ prevent us for lower $L^2$ to $L^1$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.